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znimon
 3 years ago
Solve the nonlinear inequality. Express the solution in interval notation. x^4 is greater than x^2
znimon
 3 years ago
Solve the nonlinear inequality. Express the solution in interval notation. x^4 is greater than x^2

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znimon
 3 years ago
Best ResponseYou've already chosen the best response.0I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1\[x ^{4}>x ^{2}\] \[x ^{4}x ^{2}>0\] \[x ^{2}(x ^{2}1)>0\] \[x ^{2}>0 or x ^{2}>1\] x>0 or x<1 or x>1 then x<1 and x>0 will be the interval...

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0\(x^2(x^21)>0\) divide it by \(x^2\): \((x1)(x+1)>0\) \(x\in(\infty;1)\bigcup(1;\infty)\)

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0why is it greater than 0

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1@znimon understood the process?

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0almost, what about after x^2(x^21) is greater than 0

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0\(x^2\) is always bigger than 0. It equals 0 when \(x=0\). But it doesn't satisfy the inequality.

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1took x^2 common in both the terms....

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1nops.. while splitting the inequality x has to be >0

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0will you break it down after the x^2(x^21) is greater than 0

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1yep.... both the split terms are compared to the inequalities...

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0I understand how it works so I need to find all the solutions for x that satisfy the inequality

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0thanks now I'll try some on my own. Will you recommend a problem that has the same properties?

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1try this one... \[x ^{8}x ^{2}<0\] remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1

sriramkumar
 3 years ago
Best ResponseYou've already chosen the best response.1@znimon you're welcome :)))))))))))))))))

oldrin.bataku
 3 years ago
Best ResponseYou've already chosen the best response.0\[\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 nonzero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ x\ge\sqrt1\]

oldrin.bataku
 3 years ago
Best ResponseYou've already chosen the best response.0is either 0 or nonzero **

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?

znimon
 3 years ago
Best ResponseYou've already chosen the best response.0for x^3x^2 is less than 0
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